The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is…
The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is
$\sqrt{17}$
$\frac{17}{\sqrt{15}}$
$\frac{23}{\sqrt{17}}$
$\frac{23}{\sqrt{15}}$
Solution
Slope of line $L=-\frac{b}{5}$
Slope of line $\mathrm{K}=-\frac{3}{\mathrm{c}}$
Line $L$ is parallel to line $k$.
$
\Rightarrow \frac{\mathrm{b}}{5}=\frac{3}{\mathrm{c}} \quad \Rightarrow \mathrm{bc}=15
$
$(13,32)$ is a point on $\mathrm{L}$.
$
\begin{aligned}
& \Rightarrow \frac{13}{5}+\frac{32}{b}=1 \quad \Rightarrow \frac{32}{b}=-\frac{8}{5} \\
& \Rightarrow \mathrm{b}=-20 \quad \Rightarrow \mathrm{c}=-\frac{3}{4} \\
&
\end{aligned}
$
Equation of $K: y-4 x=3$
Distance between $\mathrm{L}$ and $\mathrm{K}=\frac{|52-32+3|}{\sqrt{17}}=\frac{23}{\sqrt{17}}$